If ABCDEF is a regular hexagon then vec(...

If ABCDEF is a regular hexagon then `vec(AD)+vec(EB)+vec(FC)` equals :

2AB

3AB

4AB

Text Solution

Verified by Experts

The correct Answer is:

ABCDEF is a regular hexogen. We know from the hexagon that AD is parallel to BC
`implies AD=2BC`
Similarly, EB is parallel to FA
`implies EB = 2FA`
and FC is parallel to AB
`impliesFC =2AB`
Thus, `AD+EB +FC =2BC+2FA+2AB`
`=2(FA+AB+BC)`
`=2(FC=2(2AB)=4AB)`

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